Silly question about capacitors

[MarkE]

Wow, the lots of replies indicates that there is still a little bit of a paradoxon going on.


Personally I find it best to be compared to water. The deeper you get, the higher the pressure, in an exponentional amount, probably just like a cap: ^2.


Take a full waterbottle, make a pinhole near the bottom and see how the pressure is constantly reduced. Even atlhough there is more pressure-energy stored in the 8 to 16v, it can only flow to cap B until both have the same pressure.
Still having problems with the heat dissipation. Is gravity heat dissipation? Considering the need of flow from + to - (or in reverse, depending on popular naming) is just like gravity, I am not sure if you can explain gravity by heat dissipation. What is gravity anyway, as it is opposite to conservative models of centrifugal force.


As we see, we can utilize the energy flow with a load, but this will slow down the pressure equalisation. Now, does that mean we can lower the energy requirement to fill cap A to 16v when we allow more time to be "consumed"?

The water case is like a capacitor, but the pressure does not go up exponentially, anymore than the voltage on a capacitor goes up exponentially with stored charge.  What goes up quadratically in each case is the stored energy.

Gravity is not heat dissipation.  Nor is gravity analagous to heat dissipation.  Gravity is not opposite to centrifugal force.  Gravity is an acceleration between masses.  Gravitational fields have to the best of our ability to evaluate them always behaved in a conservative manner.  Centrifugal force, is a reaction force to centripetal force.  Neither is conservative.

[dieter]

I meant conservative in the way it is seen from science, a conservative view, contrary to  to be openminded for a new view, like, we thought we are attracted by the ground, but now the new view seems to be that we are repelled from the sky.


Not sure tho if gravity is yet officially understood by science. I mean, other than Newtons apple.

[MarkE]

I meant conservative in the way it is seen from science, a conservative view, contrary to  to be openminded for a new view, like, we thought we are attracted by the ground, but now the new view seems to be that we are repelled from the sky.


Not sure tho if gravity is yet officially understood by science. I mean, other than Newtons apple.

We float ever so gently in the sky as it buoys us slightly from the earth that so strongly attracts us.

[Farmhand]

Farmhand you can and should answer your first question by replacing the load resistor with a capacitor of like size as the input capacitor and operating the circuit until the input capacitor depletes to 50% of its starting voltage.  For your machine more than 50% will be lost, because you will have the switching losses on top of the transfer losses.

If your intent is to retain the 50% energy, then you need a third energy store to hold that energy. 

This particular problem is one of energy, so we have to stick with accounting for energy:  beginning, middle, and end.

Fine by me, I was first to mention Poynt99's very good paper on the subject and in the actual post where I said there was no loss of "Charge" I was referring to the charge only as I also said that the capacitor actually stores charge and the charge was conserved. The wasting of energy by the exercise described in the first post is not contested by me.

When i said there is no loss I was referring to the charge and I never mentioned energy in that post. I don't think you are the maker of the rules as to what I can say in any given post.

The charge is conserved but the energy is lost. The fact that the energy is lost is obvious. I was merely pointing out that if done a different way it can produce less loss. A different way means a different setup.

The actual experiment described in the first post will always produce a loss of 50% of the potential energy, but the charge will be conserved. Happy.  Sheez.

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When we release the pressure of certain pressurized gas vessels cold is produced rather than heat. True. This is the result of a loss of energy is it not ?

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Here is the post below I think might be irking you, quoted in total.

There is no loss of "charge".

Did no one look at the "Charge" in "Coulombs", If we take 2 x 10 000 uF capacitors and we have one at 16 volts the electronics assistant tells me it has 160 Millicoulombs of charge.
Now if I was to connect the two caps together and equalize the charge which results in both having 8 volts across them then the electronics assistant tells me they will both have 80 Millicoulombs of "charge" in them so added together there is still 160 Millicoulombs of charge, so no loss.. Where is the loss. The energy in a capacitor is really only "potential energy". I think the capacitor actually stores "charge", not "energy".

The voltage increase on a capacitor has a non linear effect on the "potential energy". meaning from 0 to 8 volts is less "potential energy" than from 8 volts to 16 volts, the voltage increase is the same but the potential energy increase is not.

Cheers

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I also stated to Tinsel that I may not have worded the post well. and my intention was to say that there is no assurance that the potential energy will be transformed into useful work at any specific rate of efficiency.

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[MarkE]

Fine by me, I was first to mention Poynt99's very good paper on the subject and in the actual post where I said there was no loss of "Charge" I was referring to the charge only as I also said that the capacitor actually stores charge and the charge was conserved. The wasting of energy by the exercise described in the first post is not contested by me.

When i said there is no loss I was referring to the charge and I never mentioned energy in that post. I don't think you are the maker of the rules as to what I can say in any given post.

The charge is conserved but the energy is lost. The fact that the energy is lost is obvious. I was merely pointing out that if done a different way it can produce less loss. A different way means a different setup.

The actual experiment described in the first post will always produce a loss of 50% of the potential energy, but the charge will be conserved. Happy.  Sheez.

..

When we release the pressure of certain pressurized gas vessels cold is produced rather than heat. True. This is the result of a loss of energy is it not ?

..

Here is the post below I think might be irking you, quoted in total.
..

I also stated to Tinsel that I may not have worded the post well. and my intention was to say that there is no assurance that the potential energy will be transformed into useful work at any specific rate of efficiency.

..

I haven't seen anyone contest the conservation of charge.  The only issue to cover is whether and if so how to avoid the energy loss.  As you pointed out the energy loss appears in the resistance.  If one can make the resistance a reflection of a useful load, then the internal 50% energy loss is mitigated by some amount of useful external work done.